# How to measure static mass of light?

I am learning quantum chemistry using a book.
In this book the author deduce $$p=\frac{h}{\lambda}$$ using three equations below $$E^2=m_0^2c^4+p^2c^2$$ $$E=\frac{hc}{\lambda}$$ $$m_0=0$$ But how could you say static mass of light is zero?(the third equation) I think order of the deduction is inverse.

Today I read some pages of "Matter and Light" by Louis de Broglie. In this book he deduced $$\lambda=\frac{h}{p}$$ for general corpuscle using equation of special relativity $$p=\frac{moVg}{\sqrt{{1-\beta^2}} }$$and the relationship between phase velocity and group velocity.$${Vp=\frac{c^2}{Vg}}$$ It wasn't what the author of my quantum chemistry book has written.

• @Samuel Adrian Antz Thank you for editing! Oct 2, 2022 at 2:12
• Within special relativity, it is impossible for a body of finite mass to move at the speed of light. Since light moves at the speed of light, it's mass is zero. Oct 2, 2022 at 2:31

De Broglie's equation: $$\lambda=\frac{h}{p}$$

Energy of a photon, $$E=h\nu=\frac{hc}{\lambda}=hc\frac{p}{h}=pc$$ $$\quad$$...(A)

Relativistic energy equation: $$E=\sqrt{p^2c^2+m_0^2c^4}$$ $$\quad$$...(B) ($$=mc^2$$, where m is relativistic mass: $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$), where $$E$$ is the total energy of the particle of rest mass $$m_0$$(Kinetic energy would simply be: $$mc^2-m_oc^2$$).

Now simply equate (A) and (B), for the value of $$E$$ to match from both of these equations(and it should match as both of the equations are correct), $$m_0=0$$.

Don't think that there is something sacred about photons, any particle moving with the speed of light will have total energy: $$pc$$ and zero rest mass.

It is an experimental fact that a photon is massless. To prove this theoretically in full generality, one requires to wade through some QFT. However, arguments for massive photons lead to a reductio ad absurdum even by just taking special relativity into consideration. The so-called relativistic mass $$(m: E = mc^2 \iff \vec{p} = m\vec{v})$$ of a particle of rest mass $$m_0$$ moving at a velocity $$\vec{v}$$ relative to the observer is given by, $$$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$$$ If we assume a photon to be massive, then $$m$$ ends up undefined. In the context of limits (say, $$v \to c$$), one could either say that the limit doesn't exist or $$m \to \infty \implies E \to \infty$$. None of these make much sense. However, in the same context for the massless case (say instead of $$m_0 = 0$$, we take $$m_0 \to 0$$), we end up concluding that the form for $$m$$ is indeterminate ($$0/0$$ form). This is not absurd however (unless of course the limit doesn't exist), because we simply cannot evaluate the limit in the given form. Many sources tend to describe the relativistic mass of the photon to be given by, $$$$m = \frac{E}{c^2} = \frac{h}{\lambda c}$$$$ as it also satisfies $$p = mc$$. One could perhaps think of this to be what the original limit would produce and be relieved of any nightmares. Hence one can safely conclude that for the photon, $$m_0 = 0$$.

EDIT: When it comes down to it, it is alright to replace $$\to$$ with $$=$$ for simple physical purposes like I did in the end. In truth ofc, this argument only tells us that $$m_0 \to 0$$, but that is still okay in the context of your required derivation.